Let $\overrightarrow {F} = (x+2y)e^z \hat i + (ye^z+x^2) \hat j + y^2 z \hat k$ and let $S$ be the surface $x^2+y^2 +z^2 = 1, z \geq 0.$ If $\hat n$ is the unit normal to $S$ and $$\left | \iint_S \left (\overrightarrow {\nabla} \times \overrightarrow F \right ) \cdot \hat n\ dS \right | = \alpha \pi.$$ Then find the value of $\alpha.$
If I use Gauss' divergence theorem then the integral becomes $0$ because $\operatorname {div} \left ( \operatorname {curl} \overrightarrow {F} \right ) = 0.$ On the other hand by using Stokes' theorem the integral evaluates to $2 \pi.$ Where am I doing mistake? Any help in this regard will be highly appreciated.
Thank you very much.
Gauss' divergence theorem $$\int_B{\rm div}({\bf v})\>{\rm dvol}=\int_{\partial B}{\bf v}\cdot{\bf n}\>{\rm d}\omega \tag{1} $$ applies to a 3D body $B$, its full boundary $\partial B$, and a vector field ${\bf v}$.
In your case $B$ is the upper half ball, its boundary $\partial B$ is the upper half sphere $S$ plus the unit disc $D$ in the $(x,y)$-plane, and ${\bf v}={\rm rot}(F)$. Since ${\rm div}({\rm rot}(F))\equiv0$ in the interior of $B$ the formula $(1)$ gives $$0=\int_{\partial B}{\rm rot}(F)\cdot{\bf n}\>{\rm d}\omega=\int_S{\rm rot}(F)\cdot{\bf n}\>{\rm d}\omega+\int_D{\rm rot}(F)\cdot{\bf n}\>{\rm d}\omega\ .$$ This implies that the integral of ${\rm rot}(F)$ over $S$ is equal to minus the integral over $D$, with correct orientations. The $D$-integral computes to $2\pi$, so that $\alpha=2$.